A capacitor of `2F` (practically not possible to have a capacity of `2F`) is charged by a battery of `6v`. The battery is removed and circuit is made as shown. Switch is closed at time `t=0` . Choose the correct options.

Options

A. At time `t=0` current in the circuirt is `2A`
B. At time `t=(6ln2)`second, potential difference across capacitor is `3V`
C. At time `t=(6 ln2)` second potential difference across `1Omega` resistance is `1V`
D. At time `t=(6ln2)` second, potential difference across `2Omega` resistance is `2V`

Answer

A, B, C, D

Explanation

a. At `t=0`, emf of the circuit `=PD` across the capcitor `=6V`
`:. i=6/(1+2)=2A`
Half life of the circuit
`(ln 2)tau_C(ln 2)CR=(6 ln 2)s` .
in half life time all get halved.
for example
`V_C=6/2=3V`
`i=2/1=1A`
`:. V_(1Omega)=iR=1V`
`V_(2Omega)=iR=2V`